Question

2Na2O2 + 2CO2 → 2Na2CO3 + O2
The percent yield of Na2CO3 when 7.80 g of Na2O2 is reacted with excess CO2 is 81.0 %.
What is the actual yield of Na2CO3?

If you know understand how this process works, please let me know how to do it, and the answer.
Thank you.

Answer 1

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

Answer 2

Answer:

8.58 g.

Explanation:

• To get the actual yield of a reaction, we can use the relation:

The percent yield = [actual yield]/[theoretical yield] x 100.

The percent yield = 81.0 %.

• We need to calculate the theoretical yield:

• From the balanced reaction: 2Na₂O₂ + 2CO₂ → 2Na₂CO₃ + O₂.

• It is clear that 2.0 moles of Na₂O₂ react with 2.0 moles of CO₂ to produce 2.0 moles of Na₂CO₃ and 1.0 mole of O₂.

• The no. of moles of reacted 7.80 g of Na₂O₂ = mass/molar mass = (7.80 g)/(77.98 g/mol) = 0.1 mol.

• We can get the no. of moles of produced Na₂CO₃ using cross multiplication:

∵ 2.0 moles of Na₂O₂ produce → 2.0 moles Na₂CO₃.

∴ 0.1 mol of Na₂O₂ produce → 0.1 mol Na₂CO₃.

• We can get the mass of 0.1 mol Na₂CO₃:

mass of 0.1 mol Na₂CO₃ = n x molar mass = (0.1 mol)(105.98 g/mol) = 10.598 g.

∵ The percent yield = [actual yield]/[theoretical yield] x 100.

∴ 81.0 % = [actual yield]/[10.598 g] x 100.

∴ [actual yield] = (81.0 %)(10.598)/(100) = 8.58 g.